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\lhead{Will Song}
\chead{Dirichlet's Theorem}
The $n$-th \textbf{cyclotomic polynomial} $\Phi_n$ is defined to be
\[ \Phi_n(X) = \prod_{\mathclap{\substack{\zeta^n = 1 \\ \zeta^j \neq 1 \\ 0 < j
< n}}} (X - \zeta), \]
where $\zeta$ runs over all the primitive $n$-th roots of unity.
$\deg \Phi_n = \varphi(n)$.
$X^n - 1 = \prod_{d \mid n} \Phi_d(X)$.
This is a simple exercise in root counting. Recall that $X^n - 1 =
\prod_{\zeta^n = 1} (X - \zeta)$. For each $\zeta$ such that $\zeta^n = 1$, we
can define $\ord(\zeta)$ as the smallest $k$ such that $\zeta^k = 1$. Notice
that $\ord(\zeta) \mid n$. The factor $X - \zeta$ occurs in precisely
$\Phi_{\ord(\zeta)}(X)$ and nothing else.
$n = \sum_{d \mid n} \varphi(d)$.
Trivial. Just add the degrees.
$\Phi_n(X) \in \ZZ[X]$.
We proceed with induction. The base case $\Phi_1(X) = X - 1$ is clearly true.
Now suppose for all $k < n$ we have that $\Phi_k(X) \in \ZZ[X]$. Define
\[ P_n(X) = \prod_{\mathclap{\substack{d \mid n \\ d \neq n}}} \Phi_d(X). \]
Clearly $P_n$ is monic. We can then use the division algorithm to find
integer polynomials $Q(X), R(X)$ such that
\[ X^n - 1 = P_n(X) Q(X) + R(X) \]
and $\deg R < \deg P_n = n - \varphi(n)$ or $R(X) = 0$. By plugging in the roots
of $P_n$, it is easy to see that $R(X)$ must have at least $n - \varphi(n)$
roots, so if $R$ is non-zero it must have degree at least $n - \varphi(n)$,
which cannot happen. Thus $R(X) = 0$ and $X^n - 1 = P_n(X) Q(X)$, where $Q(X)
\in \ZZ[X]$, but $X^n - 1 = P_n(X) \Phi_n(X)$, which gives us $Q(X) = \Phi_n(X)$
and $\Phi_n(X) \in \ZZ[X]$.
$\Phi_n(X)$ is irreducible.
We actually don't have enough tools to show this for the general case. It is
easy to show that $\Phi_p(X + 1)$ is irreducible and thus $\Phi_p(X)$ is also
irreducible when $p$ is prime though.
Just compute $\Phi_p(X) = 1 + X + X^2 + \cdots + X^{p - 1}$ and apply Eisenstein
at $X + 1$.
We now take some inspiration from standard calculus, and state some properties
that we will not prove. Note that these hold in $\RR[X]$ and $\ZZ[X]$, but more
importantly, they are also true in $\FF_p[X]$.
$(f(x) + g(x))' = f'(x) + g'(x)$.
$(f(x) g(x))' = f'(x) g(x) + f(x) g'(x)$.
We can now prove the following statement.
Let $P$ be a polynomial over either $\RR[X]$, $\QQ[X]$, $\ZZ[X]$, or $\FF_p[X]$.
Then there exists a non-constant polynomial $m(X)$ such that $m(X)^2 \mid P(X)$
if and only if $\gcd(P(X), P'(X)) \neq 1$.
Suppose there exists such a $m(X)$ such that $m(X)^2 \mid P(X)$. Then $P(X) =
m(X)^2 Q(X)$ and $P'(X) = 2 m(X) m'(X) Q(X) + m(X)^2 Q'(X)$, so $m(X) \mid
\gcd(P(X), P'(X))$.
For the other direction, $\gcd(P(X), P'(X)) \neq 1$ implies that they share a
root $X_0$. We claim that $m(X) = X - X_0$ satisfies the double root property
for $P(X)$. Suppose not. Then $P(X) = m(X) Q(X)$ and $Q(X_0) \neq 0$. Compute
$P'(X) = m'(X) Q(X) + m(X) Q'(X)$. Note that $P'(X_0) = 0$ because $X_0$ is a
root of $P'(X)$, but $P'(X_0) = m'(X_0) Q(X_0) + m(X_0) Q'(X_0) = 1$, which is a
$X^n - 1$ only has roots of multiplicity $1$ in any of the above rings.
Let $m, n$ be two positive integers and $p$ a prime such that $p \nmid mn$. Then
\[ \gcd(\Phi_n(X), \Phi_m(X)) = 1 \]
over $\FF_p[X]$.
By the previous proposition $X^{mn} - 1$ taken in $\FF_p[X]$ has no repeated
factors. Suppose for the sake of contradiction that $\gcd(\Phi_n(X), \Phi_m(X))
= g(X) \neq 1$. Then $g(X)^2 \mid X^{mn} - 1$, which is not true. This is
because $X^{mn} - 1 = \prod_{d \mid mn} \Phi_d(X)$ and $\Phi_n, \Phi_m$ both
show up in this product.
Let $m, n$ be two positive integers and $p$ prime such that $p \nmid mn$. Then
$\Phi_n(x), \Phi_m(x)$ cannot be both divisible by $p$ for the same value of
Suppose $\Phi_n(x) \equiv 0 \pmod{p}$ and $\Phi_m(x) \equiv 0 \pmod{p}$. Then $X
- x$ is a factor of $\Phi_n(X)$ and $\Phi_m(X)$ in $\FF_p[X]$, which cannot
For a given prime $p$, we define the standard order $\ord_p(a)$ to be the least
positive integer $k$ such that $a^k \equiv 1 \pmod{p}$.
Below we give one of the most important theorems in the theory of cyclotomic
Let $p$ be a prime. Then for all positive integers $n$ and integers $a$ such
that $\gcd(n, p) = 1$, we have $p \mid \Phi_n(a) \iff \ord_p(a) = n$.
We proceed via induction on $n$. The base case $n = 1$ is trivial since
$\Phi_1(x) = x - 1$ has a root at $x \equiv 1 \pmod{p}$. Now suppose that the
hypothesis is true for all $k < n$.
Suppose $a$ satisfies $\Phi_n(a) \equiv 0 \pmod{p}$. Recall that $a^n - 1 =
\prod_{d \mid n} \Phi_d(a) \equiv 0 \pmod{p}$, so $a^n \equiv 1 \pmod{p}$. Now
suppose that $\ord_p(a) = k \neq n$. By the induction hypothesis, we know that
$\Phi_k(a) = 0$. We also have $k \mid n$, so $\gcd(k, p) = 1$ implies $p \nmid
kn$, which contradicts proposition 12. Thus $\ord_p(a) = n$.
Now suppose $\ord_p(a) = n$. Then $\prod_{d \mid n} \Phi_d(a) = a^n - 1 \equiv
0 \pmod{p}$. If for some $d < n$ we have $\Phi_d(a) \equiv 0 \pmod{p}$, we have
$\ord_p(a) = d$ by the induction hypothesis so $\prod_{\mathclap{\substack{d
\mid n \\ d \neq n}}} \Phi_d(a) \not \equiv 0 \pmod{p}$. Thus we necessarily
have $\Phi_n(a) \equiv 0 \pmod{p}$.
We can now finally prove the main theorem.
Let $a, n$ be two relatively prime positive integers. Then there exist
infinitely many primes $p$ of the form $p \equiv a \pmod{n}$.
Okay I lied. We actually can't prove that. We can only prove the special case of
$a = 1$.
For any integer $n$ there exist infinitely many primes of the form $p \equiv 1
We first prove the following lemmas.
Let $p$ be a prime not dividing $n$. If there exists an integer $a$ such that $p
\mid \Phi_n(a)$, then $p \equiv 1 \pmod{n}$.
By theorem 13, we immediately know that $\ord_p(a) = n$, but this implies $n
\mid p - 1$, or $p - 1 \equiv 0 \pmod{n}$, giving the result.
Given an integer polynomial $f$, there exist infinitely many primes $p$ such
that there exists an integer $a_p$ such that $p \mid f(a_p)$.
We can write $f(n) = n g(n) + f(0)$. If $f(0) = 0$, then the result is obvious.
Now suppose $f(0) \neq 0$.
Suppose without loss of generality that $\lim_{n \to \infty} f(n) = +\infty$.
Let $f(0) = c$. That means there exists a sufficiently large $N$ such that $k >
N \implies f(k) > |c|$. Pick any of these $k$. Compute
\[ f(c^2 k!) = c^2 k! g(c^2 k!) + c = c (c k! g(c^2 k!) + 1). \]
Then $c k! g(c^2 k!) + 1$ must be divisible by some prime. It is also clearly
not divisible by any prime $p \leq k$ because $k! \equiv 0 \pmod{p}$. Thus this
is divisible by some prime larger than $k$ so we can just take a sequence of
sufficiently large $k$ to get our infinite primes.
Now we can prove the corollary.
We can use lemma 17 to get infinitely many pairs $p, a_p$ such that $p \mid
\Phi_n(a_p)$, and we immediately get $p \equiv 1 \pmod{n}$ by lemma 16.
The above method can be strengthened into the following proposition, but we do
not have the tools necessary for proving this, so we will just leave it at that.
For all integers $n$ and integers $a$ such that $a^2 \equiv 1 \pmod{n}$, there
are infinitely many primes $p$ of the form $p \equiv a \pmod{n}$.
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